On mars, an angry Mark Watney hurls a potato in frustration. Realizing he needs

Question

On mars, an angry Mark Watney hurls a potato in frustration. Realizing he needs to eat, he decides he should look for his wayward potato. He estimates that he threw the potato due north at an angle of 30° with the horizontal, with an initial speed of 15 m/s, and from a height of 2 m. Gravity on Mars produces a downward acceleration of 3.71 m/s^2. There was also a crosswind blowing due east that he estimates would produce an acceleration of 7 m/s^2 on the potato. What are the coordinates of the point where he should begin looking for his potato?

Explanation / Answer

Hi,

This problem is very similar to a classic parabolic movement. However, if we use a traditional cartesian plane to describe the movement we will note that the acceleration due to the crosswind cannot be placed over that space.

So it would be better to use three axis instead. In that case we have the following:

x axis (in this case oriented towards the north): there should be no acceleration in this axis.

y axis (in this case points above the potato): in this axis we have the acceleration due to gravity, which is negative.

z axis (in this case points towards the east): in this axis we have the acceleration die to the crosswind, which is positive.

Even if we have a third axis, as the third one is perpendicular to the other two, we can consider that the movement in that axis is independent from the others. So, we can solve this problem with the two main axis (x and y) and then add the displacement towards the east, after all axis that dictates the end of the movement is the y axis (and its negative acceleration).

Procedure

Movement equations:

x axis: vox = v cos(30°) = 13 m/s ; x = vx t ; vox = vx

y axis: voy = v sin(30°) = 7.5 m/s ; y = yo + voy t - (1/2) gt2 ; v = voy - gt

z axis: voz = 0 ; vz = az t ; z = (1/2)*az t2

Using the condition y=0 ; we can find the time using the equations from y axis and thn we can use that time to calculate the distance travelled over the x and z axis.

(1/2) g t2 - voy t - yo = 0 :::::: (1/2)(3.71 m/s2) t2 - (7.5 m/s) t - 2 m = 0 ::::::: 1.855 t2 - 7.5 t - 2 = 0

The solutions to the previous equations are: t = -0.25 s (which we ignore) t = 4.29 s (which we take)

x = (13 m/s)*(4.29 s) = 55.77 m

z = (1/2)*(7 m/s2 )*(4.29 s)2 = 64.41 m

So, according to our coordinate system the potato should be close to the point:

P(56.77 ; 0 ; 64.41) in meters, which means:

56.77 meters to the north, 64.41 meters to the eats and 0 meters above the ground.

I hope it helps.

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